Lecture 20

Here’s a slightly better idea:

More precisely, this definition is trying to say that for any positive distance, this sequence comes within this distance of xx.

In symbols, this can be written: ϵ>0,nNs.t.anx<ϵ.\forall\epsilon>0,\quad \exists n\in\mathbb{N}\quad\mathrm{s.t.}\quad\left|a_n-x\right|<\epsilon. Before we understand why this is wrong, we ought to make sure we know what this is supposed to mean.

I like to think of it as an argument with a very dangerous and unpleasant evil opponent. The evil opponent gets to choose a (positive real) distance, and we win if the sequence gets within that distance of xx, and we lose if it doesn’t.

In order to be sure of winning, we have to know how to beat the evil opponent whatever they say.

So, in investigating how close the sequence 3,3.1,3.14,3.141,3.1415,,3,\quad 3.1,\quad 3.14,\quad 3.141,\quad 3.1415,\quad \ldots, gets to 10001000, then if the evil opponent is stupid enough to ask “does the sequence get within distance 100000100000?” we’ll win. But we can’t just hope they’ll ask that. Instead they might ask “does the sequence get within distance 0.0010.001 of 10001000?”, and then we’d lose, because it never gets anywhere near there.

On the other hand, if we’re investigating how close the sequence gets to π\pi, we win no matter what they say. If they ask “does the sequence get within distance 10000001000000 of π\pi?”, then we say “yes, 33 is within 10000001000000 of π\pi”, and win. If they ask “does the sequence get within distance 0.00010.0001 of π\pi, then we say “yes, 3.141593.14159 is within 0.00010.0001 of π\pi”, and win. If they ask a question with an even tinier positive number in, we just take more digits and use that and say “yes”. We’re happy: the evil opponent can’t beat us.

This is a much better concept. But it’s still not right.

Here’s an example of what could go wrong. Consider the sequence 1.1,2.01,1.001,2.0001,1.00001,2.000001,1.1,\quad 2.01,\quad 1.001,\quad 2.0001,\quad 1.00001,\quad 2.000001,\quad \ldots Does it converge to 11? Does it converge to 22?

It’s lousy as an address for either: if we give these instructions to the numberline’s village postal worker, they’ll get very annoyed as they keep walking from somewhere near 11 to somewhere near 22 and back again.

But according to the definition above it would “converge” to both, because it gets as close as you like to 11 and it also gets as close as you like to 22.

So we need to find some way of saying that it has to make its mind up eventually. The obvious thing to do is to say is that (for any ϵ>0\epsilon>0) it has to get within ϵ\epsilon of xx, and then stay within ϵ\epsilon of xx forever.

This leads us to our final definition:
Definition: Let xx be a real number. A sequence of real numbers a0,a1,a2,a_0, a_1, a_2,\ldots is said to converge to xx if we have ϵ>0,NNs.t.n>N,anx<ϵ.\forall \epsilon>0,\quad \exists N\in\mathbb{N}\quad\mathrm{s.t.}\quad\forall n>N,\quad \left|a_n-x\right|<\epsilon.

So that says “no matter what positive real ϵ\epsilon our evil opponent gives us, we can point out some NN, such that all the terms aN+1,aN+2,aN+3,a_{N+1}, a_{N+2}, a_{N+3}, \ldots are all within ϵ\epsilon of xx”.

That does an excellent job of making precise the concept of “gets close and stays close forever”, and it’s the right definition!

Now, suppose we ask whether the sequence 3,3.1,3.14,3.141,3.1415,3,\quad 3.1,\quad 3.14,\quad 3.141,\quad 3.1415,\quad \ldots converges to π\pi. It does, because no matter what ϵ\epsilon our evil opponent asks about, we can find some term of the sequence beyond which all terms are within ϵ\epsilon of π\pi. For example, all terms after the (N+1)(N+1)st term are within 10N10^{-N} of π\pi.

Does that converge to 10001000? No, it never comes within 11 of 10001000 (for example), so it certainly doesn’t stay within 11 of 10001000 forever.

What about the sequence a0=1.1,a1=2.01,a2=1.001,a3=2.0001,?a_0 = 1.1,\quad a_1 = 2.01,\quad a_2 = 1.001,\quad a_3 = 2.0001,\quad \ldots? Does that converge to anything?

No, it doesn’t. In particular, it doesn’t converge to 11, because while it’s sometimes close to 11, it’s also sometimes close to 22. So there is no NN where ana_n is always within 0.10.1 of 11 for all n>Nn>N: all the odd-numbered ana_n aren’t in that range.

Similarly, it doesn’t converge to 22, because while it’s sometimes close to 22, it’s sometimes close to 11. So there is no NN where ana_n is always within 0.10.1 of 22 for all n>Nn>N: all the even-numbered ana_n aren’t in that range.

So, given the difficulties we’ve had in finding the right definition, perhaps you’ll have some sympathy for the fact that it took about two centuries to sort real analysis out properly. In what remains of the course I’ll try to make you like this definition.

Convergence proofs

Now we get back to the subject of convergence.

We say that a sequence (ai)iN=a0,a1,(a_i)_{i\in\mathbb{N}} = a_0, a_1, \ldots is convergent if it converges to some xx.

Here’s a very important fact (which is only true because of all that work we put in finding a good definition):


A sequence a0,a1,a_0, a_1, \ldots cannot converge to two different real numbers xx and yy.

Before we prove this, let’s think about what a proof might look like. Here’s a picture of a numberline, to help show us what would happen if a sequence was convergent to xx and to yy:

The two bars at the bottom of that diagram were deliberately chosen not to overlap (I made each of them extend one third of the distance from xx to yy, leaving another third of the distance between them in the middle).

And since it converges to xx and yy, eventually all the terms of the sequence should be within the interval around xx, and also all of them within the interval around yy, which is a contradiction since the intervals don’t meet.

Let’s do the working carefully.


We’ll prove this by contradiction. So, suppose it can: suppose that there is a sequence a0,a1,a_0, a_1, \ldots, which converges to two different real numbers xx and yy. Without loss of generality, we may take x<yx<y.

Since the sequence a0,a1,a_0,a_1,\ldots converges to xx, there is some NN such that, for all n>Nn>N, we have anx<yx3\left|a_n - x\right|<\frac{y-x}{3}.

Since the sequence a0,a1,a_0,a_1,\ldots converges to yy, there is some MM such that, for all n>Mn>M, we have any<yx3\left|a_n - y\right|<\frac{y-x}{3}.

But then, using the triangle inequality, for any nn bigger than both NN and MM, we have yx=yxyan+xan<yx3+yx3=23(yx)y-x=\left|y-x\right| {}\leq\left|y-a_n\right|+\left|x-a_n\right| {}<\frac{y-x}{3}+\frac{y-x}{3} {}=\frac{2}{3}(y-x) which is a contradiction as yxy-x is positive.

As a result, if a sequence is convergent, there is a unique real number to which it converges; we call that the limit of the sequence.

I understand you will have met the subject of convergence in MAS110.

That course is about streetfighting, and there you’re encouraged to use any techniques you have to hand without worrying too much about what it means.

This is a course about proof in mathematics, and tracing reasoning back to basic principles. If I set problems about convergence in MAS114, I need you to give a rigorous proof, with everything traced back to the definition of convergence (unless you’re told otherwise), rather than using the slightly vaguer methods and extra theorems you saw there!